If p is a prime number greater than 5, find the sum of the divisors of 15p.

So, this turns out to be very straightforward.  The problem is that it appears to be intimidating to lots of students because they don’t have a firm grasp algebraically on the concept of a prime number.  And this, to me at least, seems like the major stumbling block on all of number theory, which is why it is such a fertile source of good contest problems.  You will never have a firm algebraic grasp of a prime number (ignoring of course the rather large subject of algebraic number theory), so it suddenly seems that all of the high school training in algebra is no good here.  Nevertheless, the main way to solve problems is to actually attempt to solve them instead of throwing up your hands and bathering in despair like a souless zombie from Left 4 Dead Two.

So, how to actually tackle this problem?

Consider that a positive prime number p has as its divisors just 1 and p.  This means that the number 15p can only have divisors 1, 3, 5, 15, p, 3p, 5p and 15p.  This means that the sum of the divisors is

1 + 3 + 5 + 15 + p + 3p + 5p+ 15p = 24 + 24p=24(1+p).

In hindsight, not too bad at all.  When I’ve shown this to my students after letting them struggle for awhile, they usually say “Oh, well that’s pretty easy” and unfortunately don’t seem to learn from the problem.  I like to review the thinking process that students used to solve the problem, and see what they learn from their attempts.  Usually, the ones who are successful tried several attempts, including putting in small values for p to see if that helped with a solution.  I have also found that the successful ones stuck with it for longer than a minute, unless of course, a solution was found in less time.

This problem came from Problems and How to Solve Them Volume 1, which is a nice little manual put out by the Centre for Education in Mathematics and Computing at the University of Waterloo on how to help high school students with learning problems solving.  It is set as problem 15 of Chapter 6, Properties of Integers, Divisibility.  Over time, I will probably share a few more problems from the CEMC, mainly because they run a number of excellent contests for Canada that eventually lead to IMO-level of difficulty, but also start in the elementary grades, so there is a full range of challenges for students and teachers alike.

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